From the type of inequality:How we can find AM-GM

Algebra Level 4

If x 1 , x 2 , . . . , x n x_{1},x_{2},...,x_{n} are reals possitive number such as 2 ( x 1 + x 2 + . . . + x n ) 1 2(x_{1}+x_{2}+...+x_{n})\leq 1 Find minimum value of ( 1 x 1 ) ( 1 x 2 ) . . . ( 1 x n ) (1-x_{1})(1-x_{2})...(1-x_{n})


The answer is 0.5.

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1 solution

Mohammed Imran
Apr 3, 2020

By Weirstas's Inequality, we have ( 1 x 1 ) ( 1 x 2 ) . . . ( 1 x n ) 1 ( x 1 + x 2 + . . + x n ) = 1 1 2 = 1 2 (1-x_{1})(1-x_{2})...(1-x_{n}) \geq 1-(x_{1}+x_{2}+..+x_{n})=1-\frac{1}{2}=\boxed{\frac{1}{2}}

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